1. Binomial Question

Given $n=10$ and $p =0.05$

Find $n$ where $P(X< n)=0.8$

I have found

$P(X<1) = P(X=0) = 0.599$

$P(X<2) = P(X=0)+P(X=1) =0.914$

$n \in \mathbb{Z}^+$ ?

So $n= 2$ ? $n$ cannot be a decimal answer.

2. Maybe I'm misreading something. How did you calculate this because:

$pr(x<1)=pr(x=0)={{10}\choose{0}}(0.5)^{0}(0.5)^{10 }\approx 0.000977$

Looking at a binomial probability sums table, I can tell you the answer is not 2. I think you need to recheck your math. If you need help, let me know. Thanks.

Oops! I did misread. I thought p=0.5. My mistake.

Anyway, the answer should be 1, because it's the closest positive integer answer that doesn't cause the binomial probability to go over 0.8.

3. Originally Posted by Bushy
Given $n=10$ and $p =0.05$

Find $n$ where $P(X< n)=0.8$

[snip]
Using the same symbol (n in this case) to mean two different things in a question is dumb and is no doubt contributing to your confusion. Furthermore, the probability will never equal 0.8 for a discrete value but obviously X can only equal discrete values ....

The question can be repaired (barely) by stating it as follows:

Originally Posted by Bushy and modified by Mr F
Given $n=10$ and $p =0.05$

Find $x$ where $P(X< x)<0.8$

[snip]
Using trial and error is a reasonable approach to take but, as the previous poster has said, the answer is not x = 2. In fact, the answer to my edited question is x = 1.